Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $n = \dfrac{-3y + 9}{y^3 + 6y^2 - 27y} \div \dfrac{y - 5}{y^3 + 13y^2 + 36y} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{-3y + 9}{y^3 + 6y^2 - 27y} \times \dfrac{y^3 + 13y^2 + 36y}{y - 5} $ First factor out any common factors. $n = \dfrac{-3(y - 3)}{y(y^2 + 6y - 27)} \times \dfrac{y(y^2 + 13y + 36)}{y - 5} $ Then factor the quadratic expressions. $n = \dfrac {-3(y - 3)} {y(y + 9)(y - 3)} \times \dfrac {y(y + 9)(y + 4)} {y - 5} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac {-3(y - 3) \times y(y + 9)(y + 4) } { y(y + 9)(y - 3) \times (y - 5)} $ $n = \dfrac {-3y(y + 9)(y + 4)(y - 3)} {y(y + 9)(y - 3)(y - 5)} $ Notice that $(y + 9)$ and $(y - 3)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac {-3y\cancel{(y + 9)}(y + 4)(y - 3)} {y\cancel{(y + 9)}(y - 3)(y - 5)} $ We are dividing by $y + 9$ , so $y + 9 \neq 0$ Therefore, $y \neq -9$ $n = \dfrac {-3y\cancel{(y + 9)}(y + 4)\cancel{(y - 3)}} {y\cancel{(y + 9)}\cancel{(y - 3)}(y - 5)} $ We are dividing by $y - 3$ , so $y - 3 \neq 0$ Therefore, $y \neq 3$ $n = \dfrac {-3y(y + 4)} {y(y - 5)} $ $ n = \dfrac{-3(y + 4)}{y - 5}; y \neq -9; y \neq 3 $